SOLUTION: What are the solutions for sin(4x)+sin(x)=0, between the intervals 0 and pi

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Question 1127695: What are the solutions for sin(4x)+sin(x)=0, between the intervals 0 and pi
Answer by ikleyn(52821) About Me  (Show Source):
You can put this solution on YOUR website!
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What are the solutions for  sin(4x) + sin(x) = 0,  in the interval  [0,2pi).
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            Please pay attention on how I edited your condition. 


sin(4x) + sin(x) = 0.      (1)



Use the formula (one of the basic Trigonometry formula)

   sin(a) + sin(b) = 2%2Asin%28%28a%2Bb%29%2F2%29%2Asin%28%28a-b%29%2F2%29,

which is valid for any angles "a" and "b".



By applying it, you get from (1)


    2%2Asin%282.5x%29%2Asin%281.5x%29 = 0.


Thus you must consider two separate cases.



a)  sin(2.5x) = 0  ====>  2.5x = 0,  pi,  2pi,  3pi,  4pi,  5pi  ====>  x = 0,  pi%2F2.5,  2pi%2F2.5,  3pi%2F2.5,  4pi%2F2.5,  5pi%2F2.5 = 2pi = same as 0,

                          and after that the roots repeat cyclically.

                          This row of solutions is the same as  0,  2pi%2F5,  4pi%2F5,  6pi%2F5,  8pi%2F5  and 10pi%2F5 = 2pi = same as 0.


                          In more compact form, this family of solutions is  k%2A%282pi%2F5%29,  k = 0, 1, 2, 3, 4.



b)  sin(1.5x) = 0  ====>  1.5x = 0,  pi,  2pi,  3pi  ====>  x = 0,  pi%2F1.5,  2pi%2F1.5,  3pi%2F1.5 = 2pi = same as 0,

                          and after that the roots repeat cyclically.

                          This row of solutions is the same as  0,  2pi%2F3,  4pi%2F3,  6pi%2F3= 2pi = 0.


                          In more compact form, this family of solutions is  k%2A%282pi%2F3%29,  k = 0, 2, 4.


Answer.  There are two families of solutions:

         a)  k%2A%282pi%2F5%29,  k = 0, 1, 2, 3, 4  and  b)  k%2A%282pi%2F3%29,  k = 0, 2, 4.

Solved.